Download Modified True/False Indicate whether the sentence

Modified True/False
Indicate whether the sentence or statement is true or false. If false, change the identified word or phrase to make the
sentence or statement true.
____ 10. For two given surfaces, the coefficient of static friction is generally greater than the coefficient of kinetic
friction. _________________________
____ 11. For an object sliding along a ramp, the force of kinetic friction acting on the object increases as the angle of
inclination of the ramp increases. _________________________
____ 12. Laminar flow occurs when layers of a fluid flow smoothly over one another. _________________________
____ 13. Earth’s frame of reference is considered to be strictly an “inertial” frame of reference.
________________________________________
____ 14. A car accelerates from rest when a traffic light turns green. A cup of coffee that was sitting on the dashboard
of the car falls into the driver’s lap. The driver could rightly argue that the cup’s apparent motion was due to
its inertia. It had a tendency to stay still and the car accelerated from beneath it.
_________________________
____ 15. The sum of all the forces acting on a stationary object is the same as that acting on an object in uniform
motion. _________________________
Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
____ 20. The free-body diagram of a block being pushed up a rough ramp is best represented by
a. A
b. B
c. C
d. D
e. E
____ 22. An object sits at rest on a ramp. Which of the following free-body diagrams best represents the forces acting
on the object?
a. A
b. B
c. C
d. D
e. E
____ 44. An object sits at rest on a ramp. As the angle of inclination of the ramp increases, the object suddenly begins
to slide. Which of the following explanations best accounts for the object’s movement?
a. The coefficient of static friction has decreased sufficiently.
b. The force of gravity acting on the object has increased sufficiently.
c. The component of gravity along the ramp has increased sufficiently.
d. The friction has decreased sufficiently while the normal force has remained unchanged.
e. The normal force has increased sufficiently.
____ 46. Streamlining is a technique that designers use to
a. increase turbulent flow and decrease laminar flow
b. increase turbulent flow and increase laminar flow
c. decrease turbulent flow and increase laminar flow
d. decrease turbulent flow and decrease laminar flow
e. stop turbulent flow and stop laminar flow
____ 47. Which of the following does NOT utilize Bernoulli’s principle?
a. the lift force achieved by an airplane wing
b. the motion of a curve ball
c. the “spoiler” on the rear of an automobile
d. a Venturi “flowmeter”
e. All of the above utilize Bernoulli’s principle.
____ 48. Which of the following would be considered an “inertial” frame of reference?
a. a moving escalator
b. a car moving through a turn at a constant speed
c. an object in “free fall”
d. a car pulling away as a traffic light turns green
e. all of the above
____ 49. Which of the following frames of reference (inertial or noninertial) should NOT be grouped with the rest?
a. a satellite in Earth’s orbit
b. a child on a merry-go-round
c. a parachutist in “free fall”
d. a person standing on a moving escalator
e. a baseball as it is being hit by a bat
____ 51. The law of inertia holds
a. only for inertial frames of reference
b. only for noninertial frames of reference
c. for all frames of reference, both inertial and noninertial
d. only in a gravitational field
e. only for objects travelling with uniform motion
____ 52. The acceleration of an object sliding along a frictionless ramp that is inclined at an angle θ is
a. g cosθ
d. g
b. g sinθ
e. zero
c. g tanθ
Completion
Complete each sentence or statement.
55. The free-body diagram pictured below represents the forces acting on a toboggan sliding down a hill. The
force that has been omitted is the ____________________ force.
56. A force of 12 N acting in a direction 30° south of east is equivalent to a force of ____________________ N
acting south and a force of ____________________ N acting east.
62. For an object at rest on a ramp, the force of kinetic friction ____________________ as the angle of
inclination decreases.
63. In noninertial frames of reference, ____________________ forces are often invented to try to account for
observations.
64. A car travelling around a corner at a constant speed is considered to be a(n) ____________________ frame of
reference.
Matching
The diagram below shows two forces acting on an object. Match the components of these forces with the
magnitudes of their values.
a. 5.2 N
b. 4.4 N
____
____
____
____
65.
66.
67.
68.
c. 4.3 N
d. 2.5 N
F1X
F1Y
F2X
F2Y
Short Answer
75. For an object sliding on a ramp, what is the most appropriate way of resolving the forces? Why is this method
superior to others?
78. What is meant by “equilibrium” in the study of dynamics. Differentiate between static and dynamic
equilibrium.
82. An object sits at rest on a ramp inclined at an angle θ. As the angle of inclination increases, the object
suddenly starts moving. From a discussion of the force involved, explain why.
83. Differentiate between laminar flow and turbulent flow.
84. How does an object’s speed typically affect its aerodynamics? How can these aerodynamics be optimized?
85. Provide an example of a practical application of Bernoulli’s principle.
86. Discuss how Bernoulli’s principle is central to the design of an airplane wing.
87. Providing examples of each, differentiate between inertial and noninertial frames of reference.
88. Give an example of where one might be tempted to use a fictitious force to explain an observation within a
noninertial frame of reference. Provide the real explanation for its motion.
Problem
90. A 12.0-kg box is pushed along a horizontal surface by a 24-N force as illustrated in the diagram. The
frictional force (kinetic) acting on the object is 6.0 N.
(a) What is the acceleration of the object?
(b) Calculate the value of the normal force acting on the object.
(c) If the 12.0-kg object then runs into a 4.0-kg object that increases the overall friction by 3.0 N, what is the
new acceleration?
(d) What force does the 4.0-kg object exert on the 12.0-kg object when the two are moving together?
92. An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as
pictured in the diagram. There is a frictional force of 60 N acting between the wall and the 40-kg package.
The angle the cable makes to the vertical is 15°. Include an appropriate free-body diagram for each of the
following questions. (Assume two significant digits.)
(a) Calculate the force the winch must exert on the cable to slide the packages at a constant speed up the wall.
(b) If the packages are lowered with an acceleration 1.0 m/s2, what is the tension in the cable connecting the
two packages?
(c) If the packages are raised with an acceleration of 1.0 m/s2, what is the tension in the rope connecting the
two packages?
93. A pulley device is used to hurl projectiles from a ramp (µk = 0.26) as illustrated in the diagram. The 5.0-kg
mass is accelerated from rest at the bottom of the 4.0 m long ramp by a falling 20.0-kg mass suspended over a
frictionless pulley. Just as the 5.0-kg mass reaches the top of the ramp, it detaches from the rope (neglect the
mass of the rope) and becomes projected from the ramp.
(a) Determine the acceleration of the 5.0-kg mass along the ramp. (Provide free-body diagrams for both
masses.)
(b) Determine the tension in the rope during the acceleration of the 5.0-kg mass along the ramp.
(c) Determine the speed of projection of the 5.0-kg mass from the top of the ramp.
(d) Determine the horizontal range of the 5.0-kg mass from the base of the ramp.
94. A rope tow is used to move snow skiers along a 40º incline. The tow is designed to simultaneously carry 50
skiers of average mass 80.0 kg. The skiers simply stand on their skis and hold onto the rope as it pulls them
up the hill. The coefficient of kinetic friction between the skis and snow is 0.12.
(a) What total force must the winch at the top of the hill exert to pull a full complement of skiers up the hill at
a constant speed? Provide an appropriate free-body diagram.
(b) If the hill is 800.0 m long and a skier starts from rest at the top and heads straight down with maximum
acceleration for the first 3.0 s, and then continues at a constant speed until she reaches the bottom, how long
will the entire descent take? Include a new free-body diagram.
95. Crates of mass 50.0 kg must be hoisted onto a platform 8.0 m above the ground. A person exerts 600.0 N of
force on a rope that goes up and over a pulley suspended from the ceiling. The other end of the rope is
attached to the 50.0-kg mass.
(a) How long will it take the person to lift a crate from the ground to the platform while exerting maximum
force?
(b) If a 45.0-kg person grabs the free end of the rope when the crate is next to the platform in its elevated
position, what will the acceleration of the crate be as it falls?
(c) What is the tension in the rope in this case?
96. A 5.0-kg box accelerates across the floor at 1.2 m/s2 when pulled by a force of 12 N acting horizontally. The
same box is then pulled by the same force acting at an elevation of 30.0°.
(a) Draw the free-body diagrams for both situations.
(b) Determine the coefficient of sliding friction.
(c) Calculate the acceleration of the block if the force is directed 30.0° above the horizontal.
97. A piece of ice (m = 500.0 g) slides down the slope of a roof inclined at 50.0°. It starts from rest and slides 8.0
m along the roof, sliding off the edge at a height of 4.0 m above the level ground. The coefficient of kinetic
friction is 0.14.
(a) Draw a free-body diagram of the ice as it is sliding along the roof.
(b) With what speed does it leave the roof?
(c) How far away from the foot of the building does the ice land?
98. Two blocks are connected by a “massless” string over a “frictionless” pulley as shown in the diagram.
(a) Determine the acceleration of the blocks.
(b) Calculate the tension in the string .
(c) If the string broke, for what minimum value of the coefficient of static friction would the 2.0-kg block not
begin to slide?
100. Two masses, 4.0 kg and 6.0 kg, are connected by a “massless” rope over a “frictionless” pulley as pictured in
the diagram. The ramp is inclined at 30.0º and the coefficient of kinetic friction on the ramp is 0.18.
(a) Draw free-body diagrams of both masses.
(b) Determine the acceleration of the system once it begins to slide.
(c) Determine the tension in the rope.
(d) If the rope breaks when the 4.0-kg mass is 3.0 m from the bottom of the ramp, how long will it take for the
mass to slide all the way down? Include a new free-body diagram and assume the sliding mass starts from
rest.
101. Crates of mass 80.0 kg are loaded onto a truck, a vertical lift of 1.3 m. To make the work easier, a 2.6 m long
ramp is used and the crates are pushed along the ramp from the ground onto the truck. The coefficient of
friction between the crates and the ramp is 0.56.
(a) Determine the minimum force required to push a crate along the ramp. Include an appropriate free-body
diagram.
(b) If a crate is let go from rest at the top of the ramp and begins to slide, how long will it take to reach the
bottom of the ramp? Include a new free-body diagram.
(c) What minimum value for the coefficient of static friction will prevent the crate from beginning to slide if it
is let go at the top of the ramp?
102. A boy pulls a toy train (consisting of an engine and a caboose) along a rough floor, exerting 2.00 N of force as
indicated in the diagram. A frictional force of 0.60 N acts on the engine and a frictional force of 0.40 N acts
on the caboose.
(a) Draw free-body diagrams of both the engine and caboose.
(b) Determine the acceleration of the entire train.
(c) Calculate the tension in the string between the engine and the caboose.
Answer Section
MODIFIED TRUE/FALSE
10. ANS: T
11. ANS: F, decreases
12. ANS: T
13. ANS: F, strictly a noninertial frame of reference
14. ANS: T
15. ANS: T
MULTIPLE CHOICE
20.
22.
44.
46.
47.
48.
49.
51.
52.
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
E
C
C
C
E
A
D
A
B
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
2.1
2.1
2.4
2.4
2.4
2.5
2.5
2.5
2.3
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
FM1.01
FM1.01
FM1.05
FM1.01
FM1.01
FM1.05
FM1.05
FM1.05
FM1.01
COMPLETION
55. ANS:
LOC:
56. ANS:
LOC:
62. ANS:
LOC:
63. ANS:
LOC:
64. ANS:
LOC:
normal
FM1.01
6, 10
FM1.01
increases
FM1.01
fictitious
FM1.05
noninertial
FM1.05
REF: K/U
OBJ: 2.1
REF: K/U
OBJ: 2.1
REF: K/U
OBJ: 2.4
REF: K/U
OBJ: 2.5
REF: K/U
OBJ: 2.5
OBJ:
OBJ:
OBJ:
OBJ:
LOC:
LOC:
LOC:
LOC:
MATCHING
65.
66.
67.
68.
ANS:
ANS:
ANS:
ANS:
A
B
C
D
SHORT ANSWER
75. ANS:
REF:
REF:
REF:
REF:
K/U
K/U
K/U
K/U
2.2
2.2
2.2
2.2
FM1.01
FM1.01
FM1.01
FM1.01
For an object sliding along a ramp, it is best to resolve the forces both parallel and perpendicular to the ramp.
If the object is to accelerate, it will be parallel to the ramp, whereas there will be no acceleration
perpendicular to the ramp. The forces acting perpendicular to the ramp will be in equilibrium, whereas the
forces acting parallel to the ramp may or may not be in equilibrium.
REF: C
OBJ: 2.1
LOC: FM1.01
78. ANS:
Equilibrium is the state of “balanced” forces. If an object is in equilibrium, the forces acting on it are balanced
such that the net force acting on it is zero. According to Newton’s first law, a balanced force situation means
that the object is at rest (static equilibrium) or moving with uniform motion (dynamic equilibrium).
REF:
C
OBJ: 2.2
LOC: FM1.01
82. ANS:
The force of static friction (FS µSFN) is balanced by the component of gravity along the ramp (mgsin θ). As
well, the normal force FN is balanced by mgcos θ, the component of gravity perpendicular to the ramp. As the
angle θ increases, the value of cos θ decreases, reducing the value of the static friction force. In turn, the value
of sin θ increases making the value of the gravitational component along the ramp greater. At some point, this
gravitational component force overcomes the force of static friction and the object begins to slide.
REF: C
OBJ: 2.4
LOC: FM1.01
83. ANS:
Both terms refer to the motion of fluids. In laminar flow layers of fluid flow smoothly across one another. In
turbulent flow, these layers do not flow smoothly and produce disturbances in the fluid called eddies.
REF: C
OBJ: 2.4
LOC: FM1.01
84. ANS:
Typically, at low speeds, air passes over and around an object with relative ease (laminar flow). At greater
speeds, the flow tends to become much more turbulent. To optimize an object’s aerodynamics, streamlining is
useful.
REF: C
OBJ: 2.4
85. ANS:
- the motion of a curve ball
- the lift provided to a wing
LOC: FM1.01
REF: C
OBJ: 2.4
LOC: FM1.01
86. ANS:
The air flowing across the top of an airfoil has a greater distance to travel than air flowing across the bottom
side. As such, an area of relative high pressure will be created beneath the wing and an area of relative low
pressure will be created above the wing. This pressure difference provides lift to the wing.
REF: C
87. ANS:
OBJ: 2.4
LOC: FM1.01
An inertial frame of reference is one in which the law of inertia is valid. As such, an object will remain at rest
or in uniform motion unless acted upon by an external, unbalanced force. A car travelling with constant
velocity is an example of an inertial frame of reference. A noninertial frame of reference is one that is
accelerating. Objects in this type of frame of reference do not appear to obey the law of inertia. If the car
suddenly stopped, an object sitting in the back window would fly forward with no apparent force having acted
upon it. Its motion would seem to violate the law of inertia from its frame of reference.
REF: C
OBJ: 2.5
LOC: FM1.05
88. ANS:
A car is sitting motionless at a red light. The light turns green, the car pulls away and the driver feels her head
move backward. The driver might suggest that some force was acting on her head toward the rear of the car
and it was this force that was responsible for the motion of her head. This force does not actually exist.
Rather, the driver’s head has some inertia in its state of rest. When the car accelerates forward, her head will
have a tendency to remain stationary. The driver’s body is propelled forward by the seat of the car pushing
her, but her head tends to remain still. She thinks her head has moved backward when it is really her body that
has moved forward.
REF:
C
OBJ: 2.5
LOC: FM1.05
PROBLEM
90. ANS:
(a)
Free-body diagram:
FN acting up
Fg acting down
FA acting as illustrated
FK acting to the right
“Up” and “to the right” are the positive directions.
Horizontally:
The acceleration of the object is 1.0 m/s2.
(b)
Vertically:
The normal force is 1.3 × 102 N[up].
(c)
Free-body diagram:
FN acting up
Fg acting down
FA acting to the left
FK acting to the right
“Up” and “to the right” are the positive directions.
The acceleration of the two masses is 0.59 m/s2.
(d)
Free-body diagram:
FN acting up
Fg acting down
FA acting to the left
FK acting to the right
F acting to the right (force of 4.0 kg object on 12.0 kg object)
“Up” and “to the right” are the positive directions.
The 4.0-kg object exerts a force of 5.3 N on the 12.0-kg object.
REF: K/U
OBJ: 2.3
LOC: FM1.0
92. ANS:
(a)
Consider the system of the packages:
Free-body diagram:
FTsin 75° acting up
Fg acting down
FK acting down
At a constant speed, the acceleration and net force are both zero.
Using the sign convention that “up” is (–) and “down” is (+):
The force the winch must exert is 5.7 × 102 N [up].
(b)
Consider the 10-kg package:
Free-body diagram:
FT acting up
Fg acting down
The force of tension in the cable connecting the two packages is 88 N [up].
(c)
Consider the 10-kg package:
Free-body diagram:
FT acting up
Fg acting down
The force of tension in the cable connecting the two packages is 108 N [up].
REF: K/U
93. ANS:
(a)
For the 5.0-kg mass:
Free-body diagram:
OBJ: 2.3
LOC: FM1.01
FN acting perpendicular to ramp and up
Fg acting down
FT acting up along the ramp (this is the positive direction)
FK acting down along the ramp (this is the negative direction)
5.0 kg(a) = FT – µΚmg(cos θ) – mg(sin θ)
5.0 kg(a) = FT – 35.5 N
For the 20.0-kg mass:
Free-body diagram:
FT acting up (this is the negative direction)
Fg acting down (this is the positive direction)
20.0 kg(a) – 196 N – FT
Solving the system of equations:
a = 6.4 m/s2
The acceleration of the 5.0-kg mass along the ramp is 6.4 m/s2.
(b)
The tension in the cable is 68 N.
(c)
The speed of projection of the mass off the top of the ramp is 7.2 m/s.
(d)
Vertically: Let “up” be (–) and “down” be (+).
a = 9.8 m/s2
∆d = 6.0 m
Horizontal range:
The horizontal range for the projected mass is 9.5 m.
REF: K/U
94. ANS:
(a)
Free-body diagram:
OBJ: 2.3
LOC: FM1.01
FN acting perpendicular to the hill (upward)
Fg acting down
FA acting up along the hill (this is the negative direction)
FK acting down along the hill (this is the positive direction)
Parallel to slope:
The winch must be able to exert a force of 2.9 × 104 N to pull the skiers.
(b)
Free-body diagram:
FN acting perpendicular to the hill (upward)
Fg acting down
FK acting up along the hill (this is the negative direction)
ma = mg(sin θ) – µΚmg(cos θ)
a = 9.8 N/kg(sin 40º) – 0.12(9.8 N/kg)(cos40°)
= 5.40 m/s2
speed after 3.0 s:
v2 = v1 + a∆t
= 0.0 m/s + 5.40 m/s2(3.0 s)
v2 = 16.2 m/s
distance travelled during the period of acceleration:
distance remaining: 800.0 m – 24.3 m = 775.7 m
time to travel remaining distance:
The total time for the whole descent: 47.9 s + 3.0 s = 51 s.
REF: K/U
OBJ: 2.3
LOC: FM1.01
95. ANS:
(a)
Free-body diagram of the crate: FT acting up
Fg acting down
Let “up” be (–) and “down” be (+).
= –600 N + 50.0 kg(9.8 N/kg)
= –110 N
It will take 2.7 s to lift the crate.
(b)
For the 50.0-kg mass:
Free-body diagram:
FT acting up (this is the negative direction)
Fg acting down (this is the positive direction)
50.0 kg(a) = 50.0 kg(9.8 N/kg) – FT
For the 45.0-kg mass:
Free-body diagram:
FT acting up (this is the positive direction)
Fg acting down (this the negative direction)
45.0 kg(a) = 45.0 kg(9.8 N/kg) – FT
Solving the system of equations:
a = 0.52 m/s2
The acceleration of the crate will be 0.52 m/s2.
(c)
FT = 45.0 kg(a) + 45.0 kg (9.8 N/kg)
= 45.0 kg(0.52 m/s2) + 45.0 kg(9.8 N/kg)
FT = 4.6 × 102 N
The tension in the cable would be 4.6 × 102 N.
REF: K/U
OBJ: 2.3
LOC: FM1.01
96. ANS:
(a)
When the force is acting horizontally:
Free-body diagram:
FN acting up
Fg acting down
FA acting to the right
FK acting to the left
When the force is acting elevated:
Free-body diagram:
FN acting up
Fg acting down
FA acting 30° above the horizontal
FK acting to the left
(b)
Let “to the right” and “up” be (+).
Consider box when applied force is acting horizontally:
= 5.0 kg(1.2 m/s2) – 12 N
= –6.0 N
= mg
= 5.0 kg(9.8 N/kg)
= 49 N
The coefficient of sliding (kinetic) friction is 0.12.
(c)
When the force is acting 30º above the horizontal:
Vertically:
= –(–12 N(sin300) + 5.0 kg(9.8 N/kg))
= –43 N
Horizontally:
FK = µΚFN
= 0.12(43 N)
FK = 5.16 N
= 12 N(cos30º) – 5.16 N
= 5.23 N
The acceleration of the box is 1.0 m/s2.
REF: K/U
97. ANS:
(a)
Free-body diagram:
OBJ: 2.3
LOC: FM1.01
FN acting perpendicular to the roof (upward)
Fg acting down
FK acting up along the roof (this is the negative direction)
(b)
Parallel to the roof:
ma = mg(sin θ) – µΚmg(cos θ)
a = 9.8 N/kg(sin 50º) – (0.14)(9.8 N/kg)(cos 50º)
a = 6.62 m/s2
The ice leaves the roof at 1.0 × 101 m/s.
(c)
When the ice leaves the roof it becomes a projectile:
Vertically:
Solving the quadratic: ∆t = 0.406 s
Horizontally:
∆d = v(cos θ)∆t
= 10.3 m/s(cos 50°)(0.406 s)
∆d = 2.7 m
The ice lands 2.7 m from the base of the building.
REF:
K/U
OBJ: 2.3
LOC: FM1.01
98. ANS:
(a)
For the 0.80-kg mass:
Free-body diagram:
FN acting up
Fg acting down
FT acting to the right (this is the positive direction)
FK acting to the left (this is the negative direction)
0.80 kg(a) = FT – µKFN
0.80 kg(a) = FT – 0.14(0.80 kg)(9.8 N/kg)
0.80 kg(a) = FT – 1.10 N
For the 2.0-kg mass:
Free-body diagram:
FN acting perpendicular to the ramp (upward)
Fg acting down
FT acting up along the ramp (this is the negative direction)
FK acting up along the ramp
2.0 kg(a) = 2.0 kg(9.8 N/kg)(sin 30º) – FT – 0.14(2.0 kg)(9.8 N/kg)(cos 30º)
2.0 kg(a) = –FT + 7.42 N
Solving the system of equations: a = 2.3 m/s2
The system will accelerate at 2.3 m/s2.
(b)
FT = 0.80 kg(a) + 1.10 N
= 0.80 kg(2.26 m/s2) + 1.10 N
FT = 2.9 N
The tension in the string is 2.9 N.
(c)
If the block remains stationary:
FS = Fg sin θ
= 2.0 kg(9.8 N/kg)(sin 30°)
FS = 9.8 N
The minimum coefficient of static friction required is 0.58.
REF:
K/U
100. ANS:
(a)
For the 4.0-kg mass:
Free-body diagram:
OBJ: 2.3
LOC: FM1.01
FN acting perpendicular to the ramp (upward)
Fg acting down
FT acting up along the ramp (this is the positive direction)
FK acting down along the ramp (this is the negative direction)
For the 6.0-kg mass:
Free-body diagram:
Fg acting down (this is the positive direction)
FT acting up (this is the negative direction)
(b)
For the 4.0-kg mass:
4.0 kg(a) = FT – µΚmg(cos θ) – mg(sin θ)
4.0 kg(a) = FT – 13.5 N
For the 6.0-kg mass:
6.0 kg(a) = 58.8 N – FT
Solving the system of equations:
a = 4.5 m/s2
The acceleration of the 4.0-kg mass along the ramp is 4.5 m/s2.
(c)
FT = 4.0 kg(a) +13.5 N
= 4.0 kg(4.53 m/s2) + 13.5 N
FT = 32 N
The tension in the cable is 32 N.
(d)
For the block sliding down the ramp:
Free-body diagram:
FN acting perpendicular to the ramp (upward)
Fg acting down
FK acting up along the ramp (this is the negative direction)
ma = mg(sin θ) – µmg(cos θ)
a = 9.8 N/kg(sin 30º) – (0.18)(9.8 N/kg)(cos30º)
a = 3.37 m/s2
It would take 1.3 s to reach the bottom of the ramp.
REF: K/U
OBJ: 2.3
LOC: FM1.01
101. ANS:
(a)
For the 80.0-kg mass:
Free-body diagram:
FN acting perpendicular to the ramp (upward)
Fg acting down
FK acting down along the ramp (this is the positive direction)
FA acting up along the ramp (this is the negative direction)
Angle of inclination of the ramp:
θ = sin–1
= 30º
Parallel to the ramp:
= –µΚmg(cos θ) – mg(sin θ)
= –0.56(80.0 kg)(9.8 N/kg)(cos 30º) – 80.0 kg(9.8 N/kg)(sin 30º)
= –11.8 N
The minimum force required to push is 11.8 N up the ramp.
(b)
Free-body diagram:
FN acting perpendicular to the ramp (upward)
Fg acting down
FK acting up along the ramp (this is the negative direction)
ma = mg(sin θ) – µmg(cos θ)
a = 9.8 N/kg(sin 30º) – (0.56)(9.8 N/kg)(cos 30º)
a = 0.147 m/s2
It would take 5.9 S to reach the bottom of the ramp.
(c)
If the object does not slide:
FS = Fg sin θ
= 80.0 kg(9.8 N/kg)(sin 30º)
FS = 392 N
The minimum coefficient of static friction required is 0.58.
REF: K/U
102. ANS:
(a)
For the engine:
Free-body diagram:
For the caboose:
Free-body diagram:
OBJ: 2.3
LOC: FM1.01
FN acting up
Fg acting down
FK acting to the left
FA acting as indicated
FT acting to the left (force of caboose on engine)
FN acting up
Fg acting down
FK acting to the left
FT acting to the right (force of engine on caboose)
Let “to the right” and “upward” be (+).
(b)
Considering the entire train:
The train will accelerate at 2.1 m/s2.
(c)
Considering the caboose:
= 0.100 kg(2.1 m/s2) – (–0.40 N)
= 0.61 N
The tension in the string joining the engine and caboose is 0.61 N.
REF:
K/U
OBJ: 2.3
LOC: FM1.01